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1/(-1+2*n)
Limit of the function/ 1/(-1+2*n)

Limit of the function 1/(-1+2*n)

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The solution

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        1    
 lim --------
n->oo-1 + 2*n
lim⁡n→∞12n−1\lim_{n \to \infty} \frac{1}{2 n - 1} 
Limit(1/(-1 + 2*n), n, oo, dir='-')
Detail solution
Let's take the limit
lim⁡n→∞12n−1\lim_{n \to \infty} \frac{1}{2 n - 1}
Let's divide numerator and denominator by n:
lim⁡n→∞12n−1\lim_{n \to \infty} \frac{1}{2 n - 1} =
lim⁡n→∞(1n(2−1n))\lim_{n \to \infty}\left(\frac{1}{n \left(2 - \frac{1}{n}\right)}\right)
Do Replacement
u=1nu = \frac{1}{n}
then
lim⁡n→∞(1n(2−1n))=lim⁡u→0+(u2−u)\lim_{n \to \infty}\left(\frac{1}{n \left(2 - \frac{1}{n}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u}{2 - u}\right)
=
02−0=0\frac{0}{2 - 0} = 0

The final answer:
lim⁡n→∞12n−1=0\lim_{n \to \infty} \frac{1}{2 n - 1} = 0
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
02468-8-6-4-2-1010200-100
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Rapid solution [src] 
0
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Expand and simplify
Other limits n→0, -oo, +oo, 1
lim⁡n→∞12n−1=0\lim_{n \to \infty} \frac{1}{2 n - 1} = 0
lim⁡n→0−12n−1=−1\lim_{n \to 0^-} \frac{1}{2 n - 1} = -1
More at n→0 from the left
lim⁡n→0+12n−1=−1\lim_{n \to 0^+} \frac{1}{2 n - 1} = -1
More at n→0 from the right
lim⁡n→1−12n−1=1\lim_{n \to 1^-} \frac{1}{2 n - 1} = 1
More at n→1 from the left
lim⁡n→1+12n−1=1\lim_{n \to 1^+} \frac{1}{2 n - 1} = 1
More at n→1 from the right
lim⁡n→−∞12n−1=0\lim_{n \to -\infty} \frac{1}{2 n - 1} = 0
More at n→-oo
The graph
Limit of the function 1/(-1+2*n)
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